The Stable Roommates Problem and Some Extensions David
The Stable Roommates Problem and Some Extensions David Manlove University of Glasgow Department of Computing Science Supported by EPSRC grant GR/M 13329 and Nuffield Foundation award NUF-NAL-02 L 4 #0
Stable Roommates (SR): definition Input: 2 n persons; each person ranks all 2 n-1 other persons in strict order Output: a stable matching Definitions A matching is a set of n disjoint pairs of persons A blocking pair of a matching M is a pair of persons {p, q} M such that: • p prefers q to his partner in M, and • q prefers p to his partner in M A matching is stable if it admits no blocking pair L 4 #1
Stable Roommates (SR): definition Input: 2 n persons; each person ranks all 2 n-1 other persons in strict order Output: a stable matching Definitions A matching is a set of n disjoint pairs of persons A blocking pair of a matching M is a pair of persons {p, q} M such that: • p prefers q to his partner in M, and • q prefers p to his partner in M A matching is stable if it admits no blocking pair Example SR instance I 1: 2: 4 3 1 3: 2 1 4 4: 1 3 2 1: 3 2 4 L 4 #2
Stable Roommates (SR): definition Input: 2 n persons; each person ranks all 2 n-1 other persons in strict order Output: a stable matching Definitions A matching is a set of n disjoint pairs of persons A blocking pair of a matching M is a pair of persons {p, q} M such that: • p prefers q to his partner in M, and • q prefers p to his partner in M A matching is stable if it admits no blocking pair Example SR instance I 1: 2: 4 3 1 3: 2 1 4 4: 1 3 2 1: 3 2 4 L 4 #3
Stable Roommates (SR): definition Input: 2 n persons; each person ranks all 2 n-1 other persons in strict order Output: a stable matching Definitions A matching is a set of n disjoint pairs of persons A blocking pair of a matching M is a pair of persons {p, q} M such that: • p prefers q to his partner in M, and • q prefers p to his partner in M A matching is stable if it admits no blocking pair Example SR instance I 1: 2: 4 3 1 3: 2 1 4 4: 1 3 2 1: 3 2 4 The matching is not stable as {1, 3} blocks. L 4 #4
Stable matching in I 1: 2: 3: 4: 3 4 2 1 2 3 1 3 4 1 4 2 L 4 #5
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 4 2 1 3 1 2 3 1 3 2 3 1 2 4 1 4 2 4 4 4 3 L 4 #6
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 4 2 1 3 1 2 3 1 3 2 3 1 2 4 1 4 2 4 4 4 3 L 4 #7
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 4 2 1 3 1 2 3 1 3 2 3 1 2 4 1 4 2 4 4 4 3 L 4 #8
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 1 2 3 4 2 1 3 1 2 3 1 3 2 3 1 2 4 1 4 2 4 4 4 3 L 4 #9
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 1 2 3 4 2 1 3 1 2 3 1 3 2 3 1 2 4 1 4 2 4 4 4 3 L 4 #10
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 1 2 1 2 3 1 2 4 4 4 3 1: 2: 3: 4: 3 1 2 4 4 4 3 3 4 2 1 2 3 1 3 4 1 4 2 L 4 #11
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 1 2 1 2 3 1 2 4 4 4 3 1: 2: 3: 4: 3 1 2 4 4 4 3 3 4 2 1 2 3 1 3 4 1 4 2 L 4 #12
Stable matching in I 1: 2: 3: 4: Example SR instance I 2: 1: 2: 3: 4: 3 1 2 1 2 3 1 2 4 4 4 3 1: 2: 3: 4: 3 1 2 4 4 4 3 3 4 2 1 2 3 1 3 4 1 4 2 The three matchings containing the pairs {1, 4}, {2, 4}, {3, 4} are blocked by the pairs {1, 2}, {2, 3}, {3, 1} respectively. instance I 2 has no stable matching. L 4 #13
Stable marriage (SM) problem SM is a special case of SR Input: n men and n women; each person ranks all n members of the opposite sex in order Output: a stable matching Definitions: A matching is a set of n disjoint (man, woman) pairs A blocking pair of a matching M is an unmatched (man, woman) pair (m, w) such that: • m prefers w to his partner in M, and • w prefers m to her partner in M A matching is stable if it admits no blocking pair • Every instance of SM admits a stable matching • Such a matching can be found by an efficient algorithm known as the Gale/Shapley algorithm L 4 #14
Reduction from SM to SR An instance I of SM can be reduced to an instance J of SR such that set of stable matchings in I = set of stable matchings in J SM instance I: U={m 1, m 2, . . . , mn} is the set of men, and W={w 1, w 2, . . . , wn} is the set of women SR instance J: People comprise men in U and women in W mi’s list in J is his list in I together with the men in U{mi} appended in arbitrary order. wi’s list in J is her list in I together with the women in W{wi} appended in arbitrary order. Proposition: Set of stable matchings in I = set of stable matchings in J Open question: can we reduce SR to SM? L 4 #15
Efficient algorithm for SR • Knuth (1976): is there an efficient algorithm for deciding whethere exists a stable matching, given an instance of SR? • Irving (1985) “An efficient algorithm for the ‘Stable Roommates’ Problem”, Journal of Algorithms, 6: 577 -595 Ø given an instance of SR, decides whether a stable matching exists; if so, finds one • Algorithm is in two phases Ø Phase 1: similar to GS algorithm for SM Ø Phase 2: elimination of “rotations” • Phase 1: Ø if p proposes to q, p becomes assigned to q, and q becomes assigned from p Ø “assigned” relation is not necessarily symmetric Ø if anyone’s preference list becomes empty, no stable matching exists L 4 #16
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 6 3 5 6 2 1 2 5 1 5 3 4 5 1 6 2 3 4 2 3 1 3 L 4 #17
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 L 4 #18
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 1 L 4 #19
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 1 L 4 #20
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 1 L 4 #21
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 6 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 1 2 L 4 #22
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 6 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 1 2 L 4 #23
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 6 5 1 6 2 5 6 5 1 3 2 3 6 1 1 4 2 3 1 3 2 L 4 #24
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 4 3 5 1 4 6 5 1 6 2 5 6 5 1 3 2 3 6 1 1 4 2 3 1 3 2 L 4 #25
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 4 6 5 2 4 1 3 2 L 4 #26
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 4 6 5 2 4 4 5/1 3 2 L 4 #27
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 5 2 4 4 5 3 2 L 4 #28
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 6 5 2 4 4 5 3 1/2 L 4 #29
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 4 5 2 4 5 3 1 L 4 #30
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 3 5 2 4 4 2 5 3 1 L 4 #31
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 assigned to / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 3 5 2 4 1 6 4 2 5 3 1 L 4 #32
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } 1: 2: 3: 4: 5: 6: 4 6 4 2 4 5 to assigned / from 6 2 5 3 3 5 1 4 5 1 6 2 6 5 1 3 2 3 6 1 1 4 2 3 6 3 5 2 4 1 6 4 2 5 3 1 L 4 #33
Phase 1 of Algorithm SR assign everyone to be free; while ((some person p has a non-empty list) and (p is not assigned to anyone)) { q : = first person on p’s list; /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } } assigned to / from 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 6 3 5 2 4 1 6 4 2 5 3 1 L 4 #34
Some notation T denotes the set of preference lists after Phase 1 f. T(x) denotes the first person on x’s list s. T(x) denotes the second person on x’s list l. T(x) denotes the last person on x’s list At the termination of Phase 1: 1. y= f. T(x) if and only if x=l. T(y) 2. the pair {x, y} is absent if and only if x prefers l. T(x) to y, or y prefers l. T(y) to x 3. no preference list is empty assigned to / from 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 6 3 5 2 4 1 6 4 2 5 3 1 L 4 #35
Stables Any set of preference lists satisfying 1, 2 and 3 above is called a stable Lemma: Suppose that, during the execution of Phase 1, no person’s list becomes empty. Let T denote the set of preference lists at this point. Then T is a stable. Lemma: Suppose that, during the execution of Phase 1, some person’s list becomes empty. Then no stable matching exists. So if somebody’s list becomes empty during Phase 1, we can immediately halt. Otherwise, proceed to Phase 2… L 4 #36
Phase 2 of the algorithm Deals with the fact that the assignment relation is not symmetric in general Ø involves rotation eliminations 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 assigned to / from 6 4 3 5 2 3 4 1 6 4 2 5 3 1 Let T be a stable, and let =(x 0, y 0), (x 1, y 1), …, (xr-1, yr-1) be a sequence of pairs such that yi=f. T(xi) and yi+1=s. T(xi) for all i (0 i r-1). Then is called a rotation exposed in T. Example: =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) y 0 = 5 = f. T(x 0) y 1 = 2 = f. T(x 1) y 1 = 2 = s. T(x 0) y 0 = 5 = s. T(x 1) L 4 #37
Phase 2 of the algorithm Deals with the fact that the assignment relation is not symmetric in general Ø involves rotation eliminations 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 assigned to / from 6 4 3 5 2 3 4 1 6 4 2 5 3 1 Let T be a stable, and let =(x 0, y 0), (x 1, y 1), …, (xr-1, yr-1) be a sequence of pairs such that yi=f. T(xi) and yi+1=s. T(xi) for all i (0 i r-1). Then is called a rotation exposed in T. Example: =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) y 0 = 5 = f. T(x 0) y 1 = 2 = f. T(x 1) y 1 = 2 = s. T(x 0) y 0 = 5 = s. T(x 1) L 4 #38
Phase 2 of the algorithm Deals with the fact that the assignment relation is not symmetric in general Ø involves rotation eliminations 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 assigned to / from 6 4 3 5 2 3 4 1 6 4 2 5 3 1 Let T be a stable, and let =(x 0, y 0), (x 1, y 1), …, (xr-1, yr-1) be a sequence of pairs such that yi=f. T(xi) and yi+1=s. T(xi) for all i (0 i r-1). Then is called a rotation exposed in T. Example: =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) y 0 = 5 = f. T(x 0) y 1 = 2 = f. T(x 1) y 1 = 2 = s. T(x 0) y 0 = 5 = s. T(x 1) L 4 #39
Phase 2 of the algorithm Deals with the fact that the assignment relation is not symmetric in general Ø involves rotation eliminations 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 assigned to / from 6 4 3 5 2 3 4 1 6 4 2 5 3 1 Let T be a stable, and let =(x 0, y 0), (x 1, y 1), …, (xr-1, yr-1) be a sequence of pairs such that yi=f. T(xi) and yi+1=s. T(xi) for all i (0 i r-1). Then is called a rotation exposed in T. Example: =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) y 0 = 5 = f. T(x 0) y 1 = 2 = f. T(x 1) y 1 = 2 = s. T(x 0) y 0 = 5 = s. T(x 1) L 4 #40
Phase 2 of the algorithm Deals with the fact that the assignment relation is not symmetric in general Ø involves rotation eliminations 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 assigned to / from 6 4 3 5 2 3 4 1 6 4 2 5 3 1 Let T be a stable, and let =(x 0, y 0), (x 1, y 1), …, (xr-1, yr-1) be a sequence of pairs such that yi=f. T(xi) and yi+1=s. T(xi) for all i (0 i r-1). Then is called a rotation exposed in T. Example: =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) y 0 = 5 = f. T(x 0) y 1 = 2 = f. T(x 1) y 1 = 2 = s. T(x 0) y 0 = 5 = s. T(x 1) L 4 #41
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 assigned to / from 6 6 3 4 5 2 2 5 4 3 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: L 4 #42
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 6 2: 3 3: 5 4: 2 5: 4 6: 1 5 2 4 3 assigned to / from 6 6 3 4 2 2 5 4 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: y 0 rejects x 0 L 4 #43
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 6 2: 3 3: 5 4: 2 5: 4 6: 1 5 2 4 3 assigned to / from 6 6 3 3/4 2 2 2 5 4 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: y 0 rejects x 0 , so that x 0 proposes to y 1 L 4 #44
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 6 2: 3 5 3: 5 2 4: 2 5 5: 4 2 6: 1 4 3 assigned to / from 6 6 3 3 2 2 5 4 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: y 0 rejects x 0 , so that x 0 proposes to y 1 , then y 1 rejects x 1 L 4 #45
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 6 2: 3 5 3: 5 2 4: 2 5 5: 4 2 6: 1 4 3 assigned to / from 6 6 3 3 2 2 5 5 4 4 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: y 0 rejects x 0 , so that x 0 proposes to y 1 , then y 1 rejects x 1 , so that x 1 proposes to y 0 L 4 #46
Some intuition behind rotations The rotation =(x 0, y 0), (x 1, y 1) = (3, 5), (4, 2) is exposed in T 1: 6 2: 3 5 3: 5 2 4: 2 5 5: 4 2 6: 1 4 3 assigned to / from 6 6 3 3 2 2 5 5 4 4 1 1 x 0 has a preference list of length >1, and so is not assigned to / from the same person The effect of is equivalent to supposing that: y 0 rejects x 0 , so that x 0 proposes to y 1 , then y 1 rejects x 1 , so that x 1 proposes to y 0 This is called the elimination of We denote by T / the preference lists remaining after elimination of L 4 #47
This is T / : assigned to / from 1: 2: 3: 4: 5 5: 6: 6 3 2 5 6 3 2 4 1 6 3 2 5 4 1 It corresponds to the matching {1, 6}, {2, 3}, {4, 5} Lemma: If T is a stable with containing a list with at least two entries, then there is a rotation exposed in T Lemma: If T is a stable, and T / contains no empty lists, then T / is a stable Lemma: If there is a stable matching contained in T and is a rotation exposed in T, then there L 4 #48 is a stable matching contained in T /
Phase 2 of Algorithm SR T : = preference lists generated by Phase 1; while ((some list in T is of length >1) and (no list in T is empty)) { find a rotation exposed in T; T : = T / ; } if (some list in T is empty) output “no stable matching exists”; else output T; /* which is a stable matching */ Algorithm found stable matching {1, 6}, {2, 3}, {4, 5} 3: 4: 5: 6: 1: 2: 4 2 4 5 4 6 5 6 2 1 6 3 1 5 3 4 2 5 6 1 6 2 5 1 2 3 1 3 3 4 Algorithm takes cn 2 steps for some constant c L 4 #49
Stable Roommates with Ties (SRT) • Ties are permitted in any preference list Example SRT instance: 1: (2 2: 1 3: (1 4: 3 3) 3 4) 1 4 4 2 2 • More than one stability definition is possible L 4 #50
Stable Roommates with Ties (SRT) • Ties are permitted in any preference list Example SRT instance: 1: (2 2: 1 3: (1 4: 3 3) 3 4) 1 4 4 2 2 • More than one stability definition is possible Super-stability A matching M in instance J of SRT is super-stable if there is no pair {p, q} M such that • p strictly prefers q to his partner in M or is indifferent between them • q strictly prefers p to his partner in M or is indifferent between them In any instance of SRT with no ties, super-stability is the same as normal stability Recall: an SR instance can have no stable matching L 4 #51
Stable Roommates with Ties (SRT) • Ties are permitted in any preference list Example SRT instance: 1: (2 2: 1 3: (1 4: 3 3) 3 4) 1 4 4 2 2 • More than one stability definition is possible Super-stability A matching M in instance J of SRT is super-stable if there is no pair {p, q} M such that • p strictly prefers q to his partner in M or is indifferent between them • q strictly prefers p to his partner in M or is indifferent between them In any instance of SRT with no ties, super-stability is the same as normal stability Recall: an SR instance can have no stable matching L 4 #52
Stable Roommates with Ties (SRT) • Ties are permitted in any preference list Example SRT instance: 1: (2 2: 1 3: (1 4: 3 3) 3 4) 1 4 4 2 2 • More than one stability definition is possible Super-stability A matching M in instance J of SRT is super-stable if there is no pair {p, q} M such that • p strictly prefers q to his partner in M or is indifferent between them • q strictly prefers p to his partner in M or is indifferent between them In any instance of SRT with no ties, super-stability is the same as normal stability Recall: an SR instance can have no stable matching L 4 #53
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 L 4 #54
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 5 1 1 6 5 (1 3 6 4 2 3) 1 3 L 4 #55
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 5 1 1 6 5 (1 3 6 4 2 3) 1 3 L 4 #56
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 5 1 1 6 5 (1 3 6 4 2 3 4 2 3) 1 3 L 4 #57
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 3 5 1 4 1 6 2 5 (1 3) 3 6 1 4 2 3 L 4 #58
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 5 3 2: 6 3 5 1 4 3: (5 4) 1 6 2 4: 2 6 5 (1 3) 5: 4 2 3 6 1 6: 5 1 4 2 3 L 4 #59
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 5 3 2: 6 3 5 1 4 3: (5 4) 1 6 2 4: 2 6 5 (1 3) 5: 4 2 3 6 1 6: 5 1 4 2 3 L 4 #60
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 5 3 2: 6 3 5 1 4 3: (5 4) 1 6 2 4: 2 6 5 (1 3) 5: 4 2 3 6 1 6: 5 1 4 2 3 L 4 #61
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 3 5 1 4 1 6 2 5 (1 3) 3 6 1 4 2 3 L 4 #62
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2: 6 3 3: (5 4) 4: 2 6 5: 4 2 6: 5 1 2 5 1 5 3 4 5 3 1 4 6 2 (1 3) 6 1 2 3 L 4 #63
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 2: 6 3 5 3: (5 4) 1 4: 2 6 5 5: 4 2 3 6: 5 1 4 5 3 1 4 6 2 (1 3) 6 1 2 3 L 4 #64
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 3 5 1 4 1 6 2 5 (1 3) 3 6 1 4 2 3 L 4 #65
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 5 3 2: 6 3 5 1 4 3: (5 4) 1 6 2 4: 2 6 5 (1 3) 5: 4 2 3 6 1 6: 5 1 4 2 3 L 4 #66
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2: 6 3 3: (5 4) 4: 2 6 5: 4 2 6: 5 1 2 5 1 5 3 4 5 1 6 (1 6 2 3 4 2 3) 1 3 L 4 #67
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 2 3 5 4) 1 6 5 2 3 1 4 5 1 6 (1 6 2 3 4 2 3) 1 3 L 4 #68
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 2 3 5 4) 1 6 5 2 3 1 4 5 1 6 (1 6 2 3 4 2 3) 1 3 L 4 #69
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 2 3 4) 6 2 1 5 3 5 1 4 1 6 2 5 (1 3) 3 6 1 4 2 3 L 4 #70
SRT: super-stability (cont) • Algorithm for finding a super-stable matching if one exists, given an SRT instance Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 • Algorithm is in two phases Phase 1: similar to SR case – sequence of proposals and deletions – persons become provisionally assigned to / assigned from other persons Example SRT instance: 1: (6 4) 2 5 2: 6 3 5 1 3: (5 4) 1 6 4: 2 6 5 (1 5: 4 2 3 6 6: 5 1 4 2 3 4 2 3) 1 3 L 4 #71
Phase 1 of Algorithm SRT-Super while ((some person p has a non-empty list and (p is not assigned to anyone)) { for (each q at the head of p's list) { /* p proposes to q */ assign p to q; /* q is assigned from p */ for (each strict successor r of p in q's list) { if (r is assigned to q) break the assignment; delete the pair {q, r}; } if ( 2 persons are assigned to q) { break all assignments to q; for (each r tied with p in q's list) delete the pair {q, r}; } } } L 4 #72
After Phase 1: • If every person’s list has length 1 then the lists specify a super-stable matching • If some person’s list is empty then no super-stable matching exists • If nobody’s list is empty and some person’s list has length >1 then we move on to Phase 2 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 L 4 #73
After Phase 1: • If every person’s list has length 1 then the lists specify a super-stable matching • If some person’s list is empty then no super-stable matching exists • If nobody’s list is empty and some person’s list has length >1 then we move on to Phase 2 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 Phase 2: “reject and reactivate Phase 1” – select a person p whose list is of length >1 – let p be rejected by first person on his list – reactivate Phase 1 L 4 #74
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 L 4 #75
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 L 4 #76
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 L 4 #77
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 L 4 #78
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty L 4 #79
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 L 4 #80
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 L 4 #81
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 4 3 L 4 #82
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 2 5 2 4 3 L 4 #83
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 4 5 2 2 5 4 2 3 1 L 4 #84
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 L 4 #85
• Let person 2 be rejected by person 3 1: 2: 3: 4: 5: 6: 6 3 5 2 4 1 5 4 2 5 2 3 – After Phase 1, person 3’s list is empty – Reinstate the preference lists to their status after Phase 1 – Let p reject the last person on his list – Reactivate Phase 1 • Let person 2 reject person 4 1: 2: 3: 4: 5: 6: 6 3 5 5 2 2 5 4 2 1 4 3 – All lists now have length 1 L 4 #86
1: 2: 3: 4: 5: 6: 6 3 2 5 4 1 – All lists have length 1 1: (6 2: 6 3: (5 4: 2 5: 4 6: 5 4) 3 4) 6 2 1 2 5 5 1 1 6 5 (1 3 6 4 2 3) 1 3 – {1, 6}, {2, 3}, {4, 5} is a super-stable matching L 4 #87
Phase 2 of Algorithm SRT-Super T : = preference lists generated by Phase 1; while (some person p’s list in T is of length > 1) { let q be the person assigned from p ; let r be the person assigned to p; form T 1 from T by letting p be rejected by q; form T 2 from T by letting p reject r; reactivate Phase 1 on T 1; reactivate Phase 1 on T 2; if (no list in T 1 is empty) T : = T 1; else if (no list in T 2 is empty) T : = T 2; else output “no super-stable matching exists”; } output T; /* which is a super-stable matching */ L 4 #88
Phase 2 of Algorithm SRT-Super T : = preference lists generated by Phase 1; while (some person p’s list in T is of length > 1) { let q be the person assigned from p ; let r be the person assigned to p; form T 1 from T by letting p be rejected by q; form T 2 from T by letting p reject r; reactivate Phase 1 on T 1; reactivate Phase 1 on T 2; if (no list in T 1 is empty) T : = T 1; else if (no list in T 2 is empty) T : = T 2; else output “no super-stable matching exists”; } output T; /* which is a super-stable matching */ Theorem: (1) Algorithm SRT-Super is correct L 4 #89
Phase 2 of Algorithm SRT-Super T : = preference lists generated by Phase 1; while (some person p’s list in T is of length > 1) { let q be the person assigned from p ; let r be the person assigned to p; form T 1 from T by letting p be rejected by q; form T 2 from T by letting p reject r; reactivate Phase 1 on T 1; reactivate Phase 1 on T 2; if (no list in T 1 is empty) T : = T 1; else if (no list in T 2 is empty) T : = T 2; else output “no super-stable matching exists”; } output T; /* which is a super-stable matching */ Theorem: (1) Algorithm SRT-Super is correct L 4 #90
SRT: strong stability A matching M in instance J of SRT is strongly stable if there is no pair {p, q} M such that • p strictly prefers q to his partner in M • q strictly prefers p to his partner in M or is indifferent between them Example SRT instance: 1: 4 2 2: (1 3) 3: 1 (4 4: 2 3 3 4 2) 1 The matching is strongly stable. • Does there exist an efficient algorithm to decide whether a strongly stable matching exists, given an instance of SRT? L 4 #91
SRT: weak stability A matching M in instance J of SRT is weakly stable if there is no pair {p, q} M such that • p strictly prefers q to his partner in M • q strictly prefers p to his partner in M Example SRT instance: 1: 3 2: (1 3: 2 4: 1 2 3) 1 2 4 4 4 3 The matching is weakly stable. • The problem of deciding whether an instance of SRT admits a weakly stable matching is NP-complete Ø Ronn (1990) “NP-complete Stable Matching Problems”, Journal of Algorithms, 11: 285 -304 Ø Irving and Manlove (2002) “The Stable Roommates Problem with Ties”, Journal of Algorithms, 43: 85 -105 L 4 #92
Other extensions of SR • Stable Crews Problem (SC) – Each person p can have one of two roles, e. g. (A) Captain (p. A) or (B) Navigator (p. B) 1: 2: 3: 4: 4 B 1 B 1 A 2 A 3 A 4 A 2 A 1 B 2 A 3 B 2 B 2 B 2 B 4 B 4 A 3 B 3 B 1 A 4 A 3 A 1 B 3 A E. g. person 2 prefers person 1 as a navigator to person 4 as a captain L 4 #93
Other extensions of SR • Stable Crews Problem (SC) – Each person p can have one of two roles, e. g. (A) Captain (p. A) or (B) Navigator (p. B) 1: 2: 3: 4: role B A A B 4 B 1 B 1 A 2 A 3 A 4 A 2 A 1 B 2 A 3 B 2 B 2 B 2 B 4 B 4 A 3 B 3 B 1 A 4 A 3 A 1 B 3 A E. g. person 2 prefers person 1 as a navigator to person 4 as a captain – A matching M is a disjoint set of pairs of persons with roles – each person is in exactly one pair – each pair has exactly one Captain (A) and exactly one Navigator (B) – e. g. {1 B, 3 A}, {2 A, 4 B} L 4 #94
– {XA, YB} is a blocking pair of M if (1) X prefers YB to his partner in M (2) Y prefers XA to his partner in M – A matching is stable if it admits no blocking pair E. g. {2 A, 3 B} blocks the previous matching 1: 2: 3: 4: role B A A B role 1: B 2: A 3: B 4: A 4 B 1 B 1 A 2 A 3 A 4 A 2 A 1 B 2 A 3 B 2 B 2 B 2 B 4 B 4 A 3 B 3 B 1 A 4 B 1 A 4 A 3 A 1 B 3 A – The above matching is stable L 4 #95
– An instance I of SR can be reduced to an instance J of SC such that every stable matching in I corresponds to a stable matching in J and vice versa – Some instances of SC do not admit a stable matching – There is an efficient algorithm to find a stable matching, if one exists, given an instance of SC – If crews are of size 3, the problem of deciding whether a stable matching exists is NP-complete Cechlárová and Ferková (2001), “The Stable Crews Problem”, submitted for publication L 4 #96
• Stable Fixtures Problem (SF) – Choose a schedule of fixtures based on teams having preferences over each other – Two given teams can play each other at most once – Each team i has a capacity ci , indicating the maximum number of fixtures it can play – A fixture allocation A is a set of unordered pairs of teams {i, j} such that, for each i, team i is scheduled to play at most ci fixtures – Example SF instance: team capacity 1: 2 2: 2 3: 2 4: 2 preferences 2 3 4 1 2 3 L 4 #97
• Stable Fixtures Problem (SF) – Choose a schedule of fixtures based on teams having preferences over each other – Two given teams can play each other at most once – Each team i has a capacity ci , indicating the maximum number of fixtures it can play – A fixture allocation A is a set of unordered pairs of teams {i, j} such that, for each i, team i is scheduled to play at most ci fixtures – Example SF instance: team capacity 1: 2 2: 2 3: 2 4: 2 preferences 2 3 4 1 2 3 – The above fixture allocation is: {1, 2}, {1, 4}, {2, 3}, {3, 4} L 4 #98
– A blocking pair of a fixture allocation A is a pair of teams {i , j} A such that: Ø i is assigned fewer than ci teams in A or i prefers j to at least one of its assignees and Ø j is assigned fewer than cj teams in A or j prefers i to at least one of its assignees – A fixture allocation is stable if it admits no blocking pair – Example SF instance: team capacity 1: 2 2: 2 3: 2 4: 2 preferences 2 3 4 1 2 3 – The above fixture allocation is not stable as {1, 3} is a blocking pair L 4 #99
– Example SF instance: team capacity 1: 2 2: 2 3: 2 4: 2 preferences 2 3 4 1 2 3 – The above fixture allocation is stable – SR is a special case of SF (in which each team has capacity 1) – There is an efficient algorithm to find a stable fixtures allocation, if one exists, given an instance of SF Ø Irving and Scott (2002), “The Stable Fixtures Problem”, submitted for publication • Exchange-Stable Roommates Problem (ESR) • Coalition Formation Games L 4 #100
Summary • Stable Roommates problem (SR) – classical algorithm (Irving, 1985) • Stable Roommates problem with Ties (SRT) – super-stability (Irving and Manlove, 2002) – strong stability – weak stability (Ronn, 1990) • Stable Crews problem (SC) – (Cechlarova and Ferkova, 2001) • Stable Fixtures problem (SF) – (Irving and Scott, 2002) Open questions: • SRT under super-stability – rotations? • SRT under strong stability • SC with Ties • SF with Ties L 4 #101
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